Prove that $1+b+(1+c)+1/c+1+a \ge 3$ if $a, b,$ and $c$ are positive real numbers. Let $a, b, c$ be positive real numbers. prove that
$$
\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}\ge\frac{3}{1+abc},
$$
and that equality occurs if and only if $a = b = c = 1$.
What I tried:
$1$st part: I tried a brute force approach where I make a common denominator for all $4$ fractions, but the number get real big real fast.
 A: AM-GM helps:
$$\sum_{cyc}\frac{1}{a(1+b)}=\frac{1}{1+abc}\sum_{cyc}\frac{1+abc}{a(1+b)}=$$
$$=\frac{1}{1+abc}\sum_{cyc}\left(\frac{1+abc}{a(1+b)}+1\right)-\frac{3}{1+abc}=$$
$$=\frac{1}{1+abc}\sum_{cyc}\frac{1+a+ab(1+c)}{a(1+b)}-\frac{3}{1+abc}=$$
$$=\frac{1}{1+abc}\sum_{cyc}\left(\frac{1+a}{a(1+b)}+\frac{b(1+c)}{1+b}\right)-\frac{3}{1+abc}\geq$$
$$\geq\frac{1}{1+abc}\cdot6\sqrt[6]{\prod_{cyc}\frac{(1+a)}{a(1+b)}\prod_{cyc}\frac{b(1+c)}{1+b}}-\frac{3}{1+abc}=\frac{3}{1+abc}.$$
A: Hint: use the GM-HM inequality, to get to $\frac{3}{1+h}\geq \frac{3}{1+g}$, where $h$ is the harmonic mean, and $g$ is the geometric mean.
A: Let $$(a;b;c)\rightarrow (\frac{kx}{y};\frac{ky}{z};\frac{kz}{x})\Rightarrow abc=k^3$$
Then we will need to prove $$\frac{yz}{kx\left(ky+z\right)}+\frac{xz}{ky\left(kz+x\right)}+\frac{xy}{kz\left(kx+y\right)}\ge \frac{3}{k^3+1}$$
By Cauchy-Schwarz:
$$LHS=\sum_{cyc}\frac{y^2z^2}{kxyz\left(ky+z\right)}\ge \frac{\left(xy+yz+xz\right)^2}{xyz\left(x+y+z\right)k\left(k+1\right)}$$
$$\ge\frac{3xyz\left(x+y+z\right)}{xyz\left(x+y+z\right)k\left(k+1\right)}=\frac{3}{k\left(k+1\right)}$$
So we will prove $$\frac{3}{k\left(k+1\right)}\ge \frac{3}{k^3+1}\Leftrightarrow \frac{3\left(k-1\right)^2}{k\left(k+1\right)\left(k^2-k+1\right)}\ge 0$$
